Ask Question
23 June, 23:05

Given that the mass of an average linebacker at Ursinus College is 250 lbs and the radius of a pea is 0.50 cm, calculate the number of linebackers that would be required to be stuffed into the volume of a pea in order to obtain the same density as an alpha particle.

+3
Answers (1)
  1. 23 June, 23:25
    0
    1.544*10⁹ Linebackers would be required in order to obtain the same density as an alpha particle

    Step-by-step explanation:

    Assuming that the pea is spherical (with radius R = 0.5 cm = 0.005 m), then its volume is

    V = 4/3π*R³ = 4/3π*R³ = 4/3*π * (0.005 m) ³ = 5.236*10⁻⁷ m³

    the mass in that volume would be m = N*L (L = mass of linebackers=250Lbs = 113.398 Kg)

    The density of an alpha particle is ρa = 3.345*10¹⁷ kg/m³ and the density in the pea ρ will be

    ρ = m/V

    since both should be equal ρ=ρa, then

    ρa = m/V = N*L/V → N = ρa*V/L

    replacing values

    N = ρa*V/L = 3.345*10¹⁷ kg/m³ * 5.236*10⁻⁷ m³ / 113.398 Kg = 1.544*10⁹ Linebackers

    N=1.544*10⁹ Linebackers
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Given that the mass of an average linebacker at Ursinus College is 250 lbs and the radius of a pea is 0.50 cm, calculate the number of ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers